Saudi Arabia Mathematical Competitions 2012

The only such pair is $(p,q)=(37,11)$. We have $p(p-1)=(q+1)(q^2-q+1)$. Since $p$ is prime and $p>p-1$ and $q^2-q+1>q+1$, there exists an integer $m>0$ such that $$q^2-q+1=mp,(1)$$ $$p-1=m(q+1).(2)$$ Since $q^3=p^2-p+1>(p-1{)}^2=m^2(q+1{)}^2>m^2{q}^2$ we deduce that $q>m^2$. We have from (1) and (2) that $$mp\equiv 1(\mathrm{mod}q)p\equiv m+1(\mathrm{mod}q)$$ and so we obtain that $m^2+m-1$ is divisible by $q$. Since $m^2+m-1<2m^2<2q$ this means that $$q=m^2+m-1.(3)$$ Applying the substitution from (3) to (1) and (2), we obtain $$q^2-q+1=mp=m^2(q+1)+m=(q-m+1)(q+1)+m$$ which after cancellations gives $(m-3)q=0$ and therefore $m=3$. Replacing in (3) and then in (2) we obtain $(p,q)=(37,11)$, the only solution.